Extensions of the W. Mnich Problem
نویسنده
چکیده
W. Sierpinski publicized the following problem proposed by Werner Mnich in 1956: Are there three rational numbers whose sum and product are both one! In 1960, J. W. S. Cassels proved that there are no rationals that meet the Mnich condition. This paper extends the Mnich problem to fc-tuples of rationals whose sum and product are one by providing infinite solutions for all k > 3. It also provides generating forms that yield infinite solutions to the original Mnich problem in real and complex numbers, as well as providing infinite solutions for rational sums and products other than one. HISTORICAL OVERVIEW Sierpinski [6] cited a question posed by Werner Mnich as a most interesting problem, and one that at that time was unsolved. The Mnich question concerned the existence of three rational numbers whose sum and product are both one: (1) x + y + z = xyz,= 1 (x, y, z rational). Cassels [1] proved that there are no rationals that satisfy the conditions of (1). Cassels also shows that this problem was expressed by Mordell [3], in equivalent, if not exact form. Additionally, Cassels has compiled an excellent bibliography that demonstrates that the "Mnich" problem has its roots in the work of Sylvester [13] who in turn obtained some results from the 1870 work of the Reverend Father Pepin. Sierpinski [9] provides a more elementary proof of the impossibility of a weaker version of (1) , along with an excellent summary of some of the equivalent forms of the "Mnich" problem. Later, Sansone and Cassels [4] provided another proof of the impossibility of (1). 1979] EXTENSIONS OF THE W. MNICH PROBLEM 173 EXTENSIONS TO ^-TUPLES It is natural to consider the generalization of the "Mnich" problem. Do there exist ^-tuples of rational numbers such that their sums and products are both one for a given natural number "k: (2) x1 + x2 + x3 + ' ' + xk = x1x2x3 ... xk = 1, for k > 3? Sierpinski [6, p. 127] states that Andrew Schinzel has proven that there are an infinite number of solutions for every k-tuple in (2) . However, in the source cited, Trost [14] only appears to credit Schinzel with the proof of infinite &-tuples in (2) when k is of the form 4n or 4n + 1, where n is a natural number (i.e., k = 4, 5,8, 9,12, 13, etc.). Schinzel provided a general form for generating an infinite number of solutions to (2) when k = 4. He provided one case for k = 5 (viz., 1,1, 1,-1, 1 ) , but failed to demonstrate any solutions at all for (2) when the values of k are of the form 4n + 2 or 4n + 3 (viz., 6, 7,10, 11,14, 15,etc.). Explicit generating functions will now be given which prove that there are indeed an infinite number of rational ^-tuples for all k > 3 that satisfy (2). It is quite obvious that for k = 2 there are no real solutions, since xy = x + y = 1 yields the quadratic equation x # + 1 = 0 whose discriminant is -3. For k = 4, a general form was given by Schinzel [14]: (3) {n/{n 1), 1/(1 n 2 ) , (n l)/n, (1 n)/n}, n * ±1, 0, e.g. , for n = 2, 4/3 1/3 + 3/2 3/2 = (4/3)(-1/3)(3/2)(-3/2) = 1. I derived the following general generating functions for all fc-tuples greater than 4. Beyond the restrictions cited, they yield an infinite set of solutions for (2) by using any rational value of n. (4) For k= 5, \ni -l/n9 -n, 1/n, l}, n + 0, e.g., for n 2, 2 1/2 2 + 1/2 + 1 = (2)(-1/2)(-2)(1/2) (1) = 1. (5) For k = 6, {l/n(n + 1), -l/n(n + 1), (n + l ) 2 , -n, -n, -n), n + 0, -1, e.g., for n = 2, 1/12 1/12 + 9 4 2 2 = (1/12)(-1/12)(9)(-4)(-2)(-2) = 1. (6) For k = 7, {(« l) 2 , (n 1/2), (n 1/2), 1, -n, l/n(n l)(n 1/2), -l/n(n 1) (n HD), n * 0, 1, 1/2, e.g., for n = 2, 1 + 3/2 + 3/2 + 1 4 + 1/3 1/3 = (1) (3/2) (3/2) (1) (-4) (1/3) (-1/3) = 1. 174 EXTENSIONS OF THE W. MNICH PROBLEM [April Since the elements of the set U (1, -1, 1, -1) have a sum of 0 and a produce of 1, [/ forms the basis for generating all remaining explicit expressions beyond k = 7 by adjoining the elements of U onto the A:-tuple results for k = 4, 5, 6, and 7. The process is then repeated as often as is necessary as a 4-cycle. For example: (7) For k = 8, {n/n 1, n 1/n, 1 n/n9 1/1 n, 1, -1, 1, -1} = {k = 4, U), n + 0, -1, +1, or {k = 4, k = 4}. (8) For k = 9, (n, -1/n, -n, 1/n, 1, 1, -1, 1, -1} = {k = 5, £/}, n f 0. (9) For fc = 10, (fe = 6 , [/}; fo r £ = 1 1 , {k = 7, u}; fo r fc = 12 , {k = 4 , fc = 4 , k = 4} or (k = 6, fc = 6} or {k = 8, U) o r {̂c = 8, k = 4} or {A: = 7, fc = 5} or {k = 4 , J/, £/}. E t c . No claim is made here that the k-tuple form of the generating functions in (4) through (9) are unique. EXTENSIONS TO OTHER NUMBER SYSTEMS Although the conditions for generating rational roots for equation (1) have been demonstrated to be impossible, it is clear that rational roots approximating the Mnich criterion can be generated with any degree of accuracy required. Consider the example: (10) (7/3)(-5/9)(-27/35) = 1, but 7/3 5/9 27/35 = 951/845 and (11) -.726547 .540786 + 2.333333 = 1, but (-.726547)(-.5406786)(2.333333) = 0.9999999. If the solution domain for equation (1) is expanded from the rationals to the reals, then there are an infinite number of solutions of the form (a ± /F, c) which can be derived from the Mnich conditions 2a + c = 1 and (a b)o = 1. One form of the solution in reals yields the following infinite set in which a is real: (12) (a +/a 2 + l/(2a 1), a /a + l/(2a 1), 1 2a), a + 1/2, e.g., for a = 2, 2 + /1373 + 2 /137T 3 = (2 + /H/T) (2 /13/3) (-3) = 1. The generating form in (12) remains real as long as a + l/(2a 1) >. 0, which is the case provided that a > 1/2 or a <_ -.6572981. 1979] EXTENSIONS OF THE W. MNICH PROBLEM 175 This latter condition for a makes the discriminant in (12) zero and yields the only solution with two equal elements: (13) A = {#-53/216 + /13/216 + #-53/216 /13/216 + 1/6}; B = 2{l/3 #-53/216 + /13/216 #-53/216 /13/216}; A + A + B = 04) 04) (5) = 1. For values of a in the interval -.6572981 < a < 1/2, the generating form in (12) yields complex conjugate results, e.g., (14) a = 0, {±i, 1}, and a = -1/2, {(-1 ± i)/29 2}. Solutions of the Mnich problem in reals have not appeared in the literature, although Sierpinski [7, p. 176] does cite the first example in (14). Also absent from the literature is a discussion of the Mnich problem in the complex plane. Assuming that the solution for (1) is of the form (a ± i\/b, c) yields the infinite generating form with n real as follows: n , . j 1 ± i/(n n + 2) I in 2) n 2 U J ; I n ? n e . g . , for a = 4 {(1 ± ^ / 3 l ) / 4 , 1/2}. The generating form in (15) remains complex as long as (n n + 2)/(n ~ 2) >. 0, which is the case provided that n > 2 or n < -1.5213797. Note that these limits are the reciprocals of those for (12). When n is in the interval -1.5213797 <_ n < 2, (15) generates real solutions. Clearly, the generating forms (12) and (15) presented here for yielding real and complex solutions to (1) are not unique. EXTENSIONS TO OTHER CONSTANT SUMS AND PRODUCTS If the restriction in (2) that the product and sum must be equal to one is replaced by some rational number c, then a more general Mnich problem develops for rational Xi: (16) xl + x2 + x3 + • • • + xk = x1x2x$ .. . xk = o, for k >_ 2. When k = 2, the infinite generating set is of the form: (17) {x, xl(x 1)} where the product and sum = x/(x 1)? x f 1? e.g., 2 + 2 = (2) (2) = 4, and 3 + 3/2 = (3) (3/2) = 9/2. When k = 3, then x + y + z = xyz = e. If we assume that y xl(x 1) as in (17), then solving for z yields z = (x + y)/(xy 1) = x/(x x + 1). The infinite set is: n 1 0, 2, 176 EXTENSIONS OF THE W. MNICH PROBLEM [April (18) f x x 2 } x ^ 1, and the product and I x 1 "ar x + 1; sum equal x/ (x 2x + 2x 1) , e.g., 2 + 2 + 4/3 = (2)(2)(4/3) = 16/3. When k = 4, using the previous results yields the infinite set: (19) \* ^ T ^ x + l' x* x* 2x* 2x + IV X * lf e.g., 2 + 2 + 4/3 + 16/13 =(2)(2)(4/3)(16/13) = 256/39. It is obvious that this process can be generalized in a recursive way to generate infinite rational solutions for any fc-tuple. Sierpiiiski [6, p. 127] credits Schinzel with demonstrating that the elements in (16) can be restricted to integers by the following substitutions for all fc >. 2: (20) xk.1 = 2 and xk = k (fulfilled first), and X •, 9 X 2 j ^ 3 ) > k-2 1. The following table presents the results of (20).
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تاریخ انتشار 2010